Work, Energy, and Simple Machines · Lesson 9 of 13
Power
“The stairs do not care how fast you climb them, but power certainly does.”
• Distinguish power from work and energy. • Calculate average power from work and time. • Use watt and horsepower appropriately. • Calculate power from changes in kinetic or potential energy. • Compare processes that do equal work in different times.
Carry the same schoolbag up the same staircase twice: first by walking slowly and later by running. Your mass, the bag’s mass and the height are unchanged, so the gain in gravitational potential energy is the same. The work needed is therefore the same. Yet running feels more demanding because the same work is completed in less time. Power describes this difference.
Power is the rate at which work is done or energy is transferred.
| Symbol | Meaning | SI Unit |
|---|---|---|
| P | Average power | watt (W) |
| W | Work done or energy transferred | joule (J) |
| t | Time taken | second (s) |
One watt is the power when one joule of work is done in one second.
How Work and Time Affect Power
More work completed in the same time requires greater power. The same work completed in less time also requires greater power. Conversely, taking more time reduces average power even though the total work may remain unchanged. Power is therefore not another name for force, work or energy.
Problem
A weightlifter raises a 75 kg mass vertically through 2 m in 5 s. Find her average power using g = 10 m s⁻².
- 1.The work becomes gravitational potential energy: W = mgh.
- 2.Substitute: W = 75 kg × 10 m s⁻² × 2 m.
- 3.Work done = 1500 J.
- 4.Use P = W/t.
- 5.P = 1500 J ÷ 5 s = 300 J s⁻¹.
- 6.Therefore, the average power is 300 W.
Power from a Change in Kinetic Energy
When an engine accelerates a vehicle, its work appears as an increase in kinetic energy if other transfers are neglected. The work-energy theorem can therefore be used first to find work, after which division by time gives average power.
Problem
A 1000 kg car starts from rest and reaches 72 km h⁻¹ in 10 s. Find the average power associated with this increase in kinetic energy.
- 1.Convert the final speed: 72 × 5/18 = 20 m s⁻¹.
- 2.Initial speed is zero, so initial kinetic energy is zero.
- 3.Final kinetic energy = one-half × 1000 × 20² = 200000 J.
- 4.Work done by the engine on the car = change in kinetic energy = 200000 J.
- 5.Power = work ÷ time = 200000 J ÷ 10 s.
- 6.Average power = 20000 W.
Comparing Energy and Power
Problem
A crane raises the same mass to a floor twice as high and takes twice as long. Compare the energy required and the average power with the original lift.
- 1.Original energy gain is E = mgh.
- 2.At twice the height, energy gain becomes 2mgh = 2E.
- 3.Original power is P = E/t.
- 4.New power is 2E divided by 2t.
- 5.The factors of two cancel, so the new power equals P.
- 6.The second lift requires twice the energy but the same average power.
Horsepower
Power ratings for engines and pumps are sometimes expressed in horsepower. The name arose when early engine performance was compared with the work rate of horses. The SI unit remains the watt.
Problem
An engine is rated at 5 hp. Express this power in watts.
- 1.Use 1 hp = 746 W.
- 2.Multiply: 5 × 746 W = 3730 W.
- 3.Therefore, 5 hp is equivalent to 3730 W.
Equal Work Does Not Mean Equal Power
Suppose two people of equal mass climb to the same platform. Both gain the same gravitational potential energy, so both do the same work against gravity. If one takes twice as long, that person’s average power is half as large. This comparison is meaningful only when the masses and vertical heights are accounted for; a heavier person may do more work even if both take the same time.
Problem
Two equal-mass students climb the same stairs. Student A takes 8 s and Student B takes 12 s. Find the ratio of their average powers.
- 1.Both students do the same work W because mass and height are equal.
- 2.PA = W/8 and PB = W/12.
- 3.Form the ratio PA/PB = (W/8) ÷ (W/12).
- 4.Cancel W and simplify: PA/PB = 12/8 = 3/2.
- 5.Their power ratio is 3 : 2, so Student A has the greater average power.
A Reliable Method for Power Problems
First identify the energy change or work done. For vertical lifting, this is usually mgh. For acceleration from one speed to another, it is the change in kinetic energy. Only after finding work in joules should it be divided by time in seconds. This order prevents power from being confused with force and keeps units consistent.
Problem
A pump raises 200 kg of water through 6 m in 20 s. Find its ideal average power using g = 10 m s⁻².
- 1.Potential-energy gain = mgh.
- 2.Work = 200 kg × 10 m s⁻² × 6 m = 12000 J.
- 3.Time = 20 s.
- 4.Power = 12000 J ÷ 20 s = 600 W.
- 5.The value is ideal because energy losses in the pump have not been included.
Interpreting a Power Rating
A power rating states the rate at which a device can transfer energy under the specified conditions; it does not state the total energy used. A 600 W pump operating for a longer time transfers more energy than the same pump operating briefly. Rearranging P = W/t gives W = Pt, so the energy transferred is the product of power and operating time. This also explains why two devices with different power ratings can sometimes complete the same total work if they operate for different durations.
Two people may do equal work but have different powers. Always include the time interval when comparing how rapidly work is done.
Quiz
Which description best matches Power?
Which description best matches Watt?
Which term matches this description: Power is the rate at which work is done or energy is transferred.
Which term matches this description: One watt is the power when one joule of work is done in one second.
Which statement is a key takeaway from this lesson?
Practice Problems
- A machine does 2400 J of work in 8 s. Find its average power.
- Two students climb the same stairs in 10 s and 15 s. If their masses are equal, compare their powers.
- A pump transfers 18000 J in 30 s. Express its power in watts.
- Convert 2 hp into watts.
Key Takeaways
• Power measures how rapidly work is done. • P = W/t. • The SI unit is the watt, equal to one joule per second. • Equal work in less time means greater power. • Work-energy calculations can be used before calculating power.